Poj 2112(最大流)

Optimal Milking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 10223		Accepted: 3718
Case Time Limit: 1000MS

Description

FJ has moved his K (1 <= K <= 30) milking machines out into the cow pastures among the C (1 <= C <= 200) cows. A set of paths of various lengths runs among the cows and the milking machines. The milking machine locations are named by ID numbers 1..K; the cow locations are named by ID numbers K+1..K+C. 

Each milking point can "process" at most M (1 <= M <= 15) cows each day. 

Write a program to find an assignment for each cow to some milking machine so that the distance the furthest-walking cow travels is minimized (and, of course, the milking machines are not overutilized). At least one legal assignment is possible for all input data sets. Cows can traverse several paths on the way to their milking machine. 

Input

* Line 1: A single line with three space-separated integers: K, C, and M. 

* Lines 2.. ...: Each of these K+C lines of K+C space-separated integers describes the distances between pairs of various entities. The input forms a symmetric matrix. Line 2 tells the distances from milking machine 1 to each of the other entities; line 3 tells the distances from machine 2 to each of the other entities, and so on. Distances of entities directly connected by a path are positive integers no larger than 200. Entities not directly connected by a path have a distance of 0. The distance from an entity to itself (i.e., all numbers on the diagonal) is also given as 0. To keep the input lines of reasonable length, when K+C > 15, a row is broken into successive lines of 15 numbers and a potentially shorter line to finish up a row. Each new row begins on its own line. 

Output

A single line with a single integer that is the minimum possible total distance for the furthest walking cow. 

Sample Input

2 3 2
0 3 2 1 1
3 0 3 2 0
2 3 0 1 0
1 2 1 0 2
1 0 0 2 0

Sample Output

2

Source

USACO 2003 U S Open

让每头牛都能和机器匹配，牛与机器间有个距离，要使最大距离最小。二分答案建图，因为机器可以匹配多次，所以只能用最大流做。先跑一遍floyed，再二分即可。

#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 300 + 10;
const int INF = 1000000000;

int M[maxn][maxn];
struct Edge {
  int from, to, cap, flow;
};

struct Dinic {
  int n, m, s, t;
  vector<Edge> edges;    // 边数的两倍
  vector<int> G[maxn];   // 邻接表，G[i][j]表示结点i的第j条边在e数组中的序号
  bool vis[maxn];        // BFS使用
  int d[maxn];           // 从起点到i的距离
  int cur[maxn];         // 当前弧指针

  void ClearAll(int n) {
    for(int i = 0; i < n; i++) G[i].clear();
    edges.clear();
  }

  void ClearFlow() {
    for(int i = 0; i < edges.size(); i++) edges[i].flow = 0;
  }

  void AddEdge(int from, int to, int cap) {
    edges.push_back((Edge){from, to, cap, 0});
    edges.push_back((Edge){to, from, 0, 0});
    m = edges.size();
    G[from].push_back(m-2);
    G[to].push_back(m-1);
  }

  bool BFS() {
    memset(vis, 0, sizeof(vis));
    queue<int> Q;
    Q.push(s);
    vis[s] = 1;
    d[s] = 0;
    while(!Q.empty()) {
      int x = Q.front(); Q.pop();
      for(int i = 0; i < G[x].size(); i++) {
        Edge& e = edges[G[x][i]];
        if(!vis[e.to] && e.cap > e.flow) {
          vis[e.to] = 1;
          d[e.to] = d[x] + 1;
          Q.push(e.to);
        }
      }
    }
    return vis[t];
  }

  int DFS(int x, int a) {
    if(x == t || a == 0) return a;
    int flow = 0, f;
    for(int& i = cur[x]; i < G[x].size(); i++) {
      Edge& e = edges[G[x][i]];
      if(d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap-e.flow))) > 0) {
        e.flow += f;
        edges[G[x][i]^1].flow -= f;
        flow += f;
        a -= f;
        if(a == 0) break;
      }
    }
    return flow;
  }

  int Maxflow(int s, int t) {
    this->s = s; this->t = t;
    int flow = 0;
    while(BFS()) {
      memset(cur, 0, sizeof(cur));
      flow += DFS(s, INF);
    }
    return flow;
  }
};

Dinic g;

int n;
void floyed(){
    for(int k = 1;k <= n;k++){
        for(int i = 1;i <= n;i++){
            for(int j = 1;j <= n;j++){
                M[i][j] = min(M[i][k]+M[k][j],M[i][j]);
            }
        }
    }
}

int main(){
    int k,c,m;
    while(scanf("%d%d%d",&k,&c,&m) != EOF){
        n = k+c;
        for(int i = 1;i <= n;i++){
            for(int j = 1;j <= n;j++){
                scanf("%d",&M[i][j]);
                if(M[i][j] == 0 && i != j) M[i][j] = INF;
            }
        }

        floyed();
        int l = 1,r = INF;
        int ans;
        int sorce = 0,sink = n+1;
        while(l <= r){
            int mid = l+(r-l)/2;
            g.ClearAll(n+2);
            for(int i = 1;i <= k;i++) g.AddEdge(sorce,i,m);
            for(int i = k+1;i <= n;i++) g.AddEdge(i,sink,1);
            for(int i = 1;i <= k;i++){
                for(int j = k+1;j <= n;j++){
                    if(M[i][j] <= mid) g.AddEdge(i,j,1);
                }
            }
            if(g.Maxflow(sorce,sink) == c){
                ans = mid;
                r = mid-1;
            }
            else l = mid+1;
        }
        printf("%d\n",ans);
    }
    return 0;
}